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  1. Practice
  2. Tutorials
  3. 30 Days of Code
  4. Day 19: Interfaces
  5. Discussions

Day 19: Interfaces

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  • saxtorff
     + 0 comments

    No Python?! :-(

  • Bob_Roebling
     + 0 comments

    C# - Just kept it simple, I'm sure the performance geeks are screaming.

    public class Calculator : AdvancedArithmetic {
    public int divisorSum(int n) {
        int sum = 0;
        for (int i = 1; i <= n; i++) {
            if (n % i == 0) sum += i;
        }
        return sum;
    }
}

  • dillingerjames
     + 0 comments

    Here is a simple solution written in Python 3. Surprisingly, many other solutions I see don't realize you don't need to iterate though each number - only the first half.

    def divisorSum(self, n):
        if n == 1:
            return 1
        else:
            factor_sum = 1 + n 
            for i in range(2, n//2 + 1):
                if n % i == 0:
                    factor_sum += i
            return factor_sum

  • RodneyShag
     + 0 comments

    Efficient O(n^(1/2)) solution

    From my HackerRank solutions.

    Runtime: O(n^(1/2))
    Space Complexity: O(1)

    public int divisorSum(int n) {
    int sum = 0;
    int sqrt = (int) Math.sqrt(n);

    // Small optimization: if n is odd, we can't have even numbers as divisors
    int stepSize = (n % 2 == 1) ? 2 : 1;

    for (int i = 1; i <= sqrt; i += stepSize) {
        if (n % i == 0) { // if "i" is a divisor
            sum += i + n/i; // add both divisors
        }
    }

    // If sqrt is a divisor, we should only count it once
    if (sqrt * sqrt == n) {
        sum -= sqrt;
    }
    return sum;
}

  • manmohansaraswat
     + 0 comments

    Here is Mine Code in Python : 

    def divisorSum(self, n):
    sum = 0
    for i in range(1,n+1):
        if n % i == 0:
            sum += i
    return sum
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